d and f Block Elements

The element with atomic number 24 is Chromium (Cr). Chromium commonly exhibits several positive oxidation states, which are due to the loss of electrons from both the 4s and 3d orbitals.

Here are the common oxidation states for Chromium :

1. +2 (as in ${Cr}^{2+} $) : This is observed where the electron configuration is $[{Ar}] 3d^4$


2. +3 (as in ${Cr}^{3+} )$ : This is the most stable state, with the electron configuration $[ {Ar} ] 3d^3$


3. +6 (as in ${CrO}_4^{2-} $ and ${Cr}_2\text{O}_7^{2-} )$ : In these cases, chromium exhibits an oxidation state of +6.

Hence, the common positive oxidation states for Chromium are +2, +3, and +6.

Therefore, the correct option is :

Option B : +2 to +6.

(A) ${}_{58}Ce \to [Xe]4{f^2}5{d^0}6{s^2}$

In complex $C{e^{ + 4}} \to [Xe]4{f^0}5{d^0}6{s^0}$

There is no unpaired electron, so, $\mu$_m = 0 BM

(B) ${}_{64}Gd \to [Xe]4{f^7}5{d^1}6{s^2}$

In complex, ${}_{64}G{d^{2 + }} \to [Xe]4{f^7}5{d^1}$

There are 8 unpaired electrons.

$\therefore$ ${\mu _m} = \sqrt {8(8 + 2)} = \sqrt {80} = 8.94BM$

(C) ${}_{63}Eu \to [Xe]4{f^9}5{d^0}6{s^0}$

In complex ${}_{63}E{u^{ + 3}} \to [Xe]4{f^6}5{d^0}6{s^0}$

contain six unpaired electrons.

So, ${\mu _m} = \sqrt {6(6 + 2)} = \sqrt {48} BM = 6.93BM$

Hence, order of spin only magnetic moment

$B > C > A$.

Generally, due to decrease in metallic radius and increase in atomic mass density increase across the period from left to right.

Correct order is Cu > Co > Fe > Cr > Zn.

$X\buildrel \Delta \over \longrightarrow {O_2} + Y + Z$ and $Z\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{KN{O_3}}^{KOH}} Y$

$\mathop {2KMn{O_4}}\limits_{(X)} \buildrel {} \over \longrightarrow \mathop {{K_2}Mn{O_4}}\limits_{(Y)} + \mathop {Mn{O_2}}\limits_{(Z)} + {O_2}$

$\mathop {Mn{O_2}}\limits_{(Z)} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{KN{O_3}}^{KOH}} \mathop {{K_2}Mn{O_4}}\limits_{(Y)} $

$\therefore$ $X = KMn{O_4},Y = {K_2}Mn{O_4}$ and $Z = Mn{O_2}$

(A) Electronic configuration of $\mathrm{Cu}$ is

$1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1$

$\Delta_i H_1$ is low while $\Delta_i H_2$ is highest because of fully filled $d$-orbital.

(B) Electronic configuration of $\mathrm{Zn}$ is $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2$

First 2 electrons are removed from $s$-orbital while 3rd electron released from fully filled $d$-orbital, which require large amount of energy.

(C) $M$ in $\left[M(\mathrm{CO})_6\right]$ is $\mathrm{Cr}$.

(D) Ni has highest heat of atomisation in the $3 d$-transition series.

(a) Basic character of metal oxides decreases with increase in oxidation number

$\stackrel{+2}{M \mathrm{O}}, \stackrel{+3}{M_2}\mathrm{O}_3, \stackrel{+4}{M} \mathrm{O}_2, \stackrel{+5}{M_2} \mathrm{O}_5$
(Decreasing basics character)

(b) Manganese ($\mathrm{Mn}$) has maximum number of oxidation state.

Sc, V, Cr, Mn (Increasing number of oxidation states)

(c) Magnetic moment $=\sqrt{n(n+1)}$

( $n=$ number of unpaired electron)

Magnetic moment $\propto$ unpaired electron

So, the correct order of magnetic moment is

$d^5>d^4>d^3>d^1$

(d) $\begin{aligned} \mathrm{Mn}^{2+} & \rightarrow 3 d^5 \\ \mathrm{Fe}^{2+} & \rightarrow 3 d^6 \\ \mathrm{Cr}^{2+} & \rightarrow 3 d^4 \\ \mathrm{Co}^{2+} & \rightarrow 3 d^7 \end{aligned}$

$\mathrm{Mn}^{2+}$ is most stable as $d$-subshell is half filled.

The electronic configuration of $\mathrm{Cu}^{2+}$ is

$[\mathrm{Ar}] 3 d^9$

It contains one unpaired electron,

Hence, $\mathrm{CuCl}_2$ is coloured compound. In all other compounds,

$\begin{aligned} & \mathrm{CuCl} \rightarrow\left[\mathrm{Cu}^{+}\right] \Rightarrow 3 d^{10} 4 s^0 \\ & \mathrm{ScCl}_3 \rightarrow\left[\mathrm{Sc}^{3+}\right] \Rightarrow 3 p^6 4 s^2 \\ & \mathrm{TiCl}_4 \rightarrow\left[\mathrm{Ti}^{4+}\right] \Rightarrow 3 p^6 4 s^2 \end{aligned}$

No unpaired electrons are present.

For general electronic configuration of metal,

$\mathrm{Fe}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^6, 4 s^2$

Now, for $\mathrm{Fe}=[\mathrm{Ar}] 3 d^6 4 s^2$

$\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 d^6 4 s^0$ [Remove $2 e^{-}$ from $s$-orbital first]

$\Rightarrow \quad \mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 d^6 4 s^0$

$\mathrm{Fe}^{3+}$ is denoted as ferric (III) ion.

Fe(Z = 26) = [Ar]3d$^6$ 4s$^2$

Fe$^{2+}$ = [Ar]3d$^6$ 4s$^0$

Total number of unpaired electrons = 4

i.e., n = 4

Magnetic moment, $\mu=\sqrt{n(n+2)}$

$=\sqrt{4(4+2)}=\sqrt{4\times6}$

$=\sqrt{24}$

$=4.9$ BM

Hence, magnetic moment of Fe$^{2+}$ is 4.9 BM.

$\mathrm{CrO}_4^{2-}$ is a stronger oxidising agent as compared to $\mathrm{MnO}_4^{2-}$.

$\mathrm{Cr}^{+6} \rightarrow[\mathrm{Ar}] 3 d^0 4 s^0$ which is stable. So, Cr reading loses its 6 electrons to attain stable noble gas and acting as oxidising agent but in $\mathrm{Mn}^{+6}$ [electronic configuration is $[\mathrm{Ar}] 3 d^1 4 s^0$ ] one electron is difficult to remove and acquire stable form. Hence, $\mathrm{MnO}_4^{-}$ is not oxidising agent but $\mathrm{CrO}_4{ }^{2-}$ is good oxidising agent.

The given electronic configuration is [Ar] 3d$^5$ 4s$^2$. To determine the group of the periodic table to which this element belongs, we need to analyze its electron configuration, particularly the electrons in the outermost shells.

In this configuration:

• [Ar] represents the argon core, which is a stable, noble gas with 18 electrons.
• 3d$^5$ indicates there are 5 electrons in the 3d subshell.
• 4s$^2$ indicates there are 2 electrons in the 4s subshell.

Total electrons in the outer shells (valence electrons) = 5 (in 3d subshell) + 2 (in 4s subshell) = 7 valence electrons.

This pattern of electron configuration is characteristic of the elements in Group 7 of the periodic table. The 3d subshell is filling and the outer 4s subshell contains 2 electrons, typical of a transition metal in Group 7. Examples of such elements include manganese (Mn), which has the atomic number 25 and the electron configuration [Ar] 3d$^5$ 4s$^2$.

Therefore, the element with the electronic configuration [Ar] 3d$^5$ 4s$^2$ belongs to:

Option C: Seventh

To identify the incorrect statement, let's analyze each option :

Option A : "In manganate and permanganate ions, the π-bonding takes place by overlap of p-orbitals of oxygen and d-orbitals of manganese." This statement is true. In both manganate ${MnO}_4^{2-} $ and permanganate ${MnO}_4^- $ ions, the manganese atom is in a high oxidation state (+6 and +7, respectively) and forms bonds with oxygen atoms involving the overlap of p-orbitals of oxygen and d-orbitals of manganese.

Option B : "Manganate ion is green in colour and permanganate ion in purple in colour." This statement is true. Manganate ions are indeed green and permanganate ions are purple in aqueous solution.

Option C : "Manganate and permanganate ions are paramagnetic." This statement is incorrect. Both manganate and permanganate ions are diamagnetic, not paramagnetic. In these ions, manganese is in +6 and +7 oxidation states, respectively, and has no unpaired electrons.

Option D : "Manganate and permanganate ions are tetrahedral." This statement is true. Both manganate and permanganate ions have a tetrahedral geometry, with the manganese atom at the center and the four oxygen atoms surrounding it.

Therefore, the incorrect statement is option C.

$\mathrm{La}^{3+}(Z=57)$ has 54 electrons.

$\mathrm{Ti}^{3+}(Z=22)$ has 19 electrons.

$\mathrm{Lu}^{3+}(Z=71)$ has 68 electrons.

and $\mathrm{Sc}^{3+}(Z=21)$ has 18 electrons.

The species with unpaired electrons show colour in aqueous solution.

Because, $\quad \mathrm{Ti}^{3+}=[\mathrm{Ar}] \cdot 4 s^1$

Thus, has one unpaired electron.

Hence, option (b) is the correct answer.

Trans uranium elements are elements after uranium, i.e. trans uranium elements have atomic number, $Z>92$.

The highest oxidation state observed in the first row transition metals is +7 .

In first row ' $\mathrm{Mn}^{\prime}$ with outer configuration : $3 d^5, 4 s^2$ show +7 oxidation state, i.e.

It has 7 electrons in outermost orbital, so has +7 oxidation state.

Among the given series of transition metal ions, one in which all metal ions have $3 d^2, 3 p^6$ electronic configuration is :

Option (d), i.e.

$\mathrm{Ti}(Z=22) \rightarrow \mathrm{Ti}^{2+}$ (has 20 electrons)

$\mathrm{V}(\mathrm{Z}=23) \rightarrow \mathrm{V}^{3+}$ (has 20 electrons)

$\mathrm{Cr}(Z=24) \rightarrow \mathrm{Cr}^{4+}$ (has 20 electrons)

and $\mathrm{Mn}(Z=25) \rightarrow \mathrm{Mn}^{5+}$ (has 20 electrons)

Thus, all can show $3 d^2 3 p^6$ electronic configuration i.e. option (d) is the correct answer.

Interstitial compounds are formed, when small atoms are trapped inside the crystal lattice of metals.

(i) They are very hard and rigid.

(ii) They have high melting point, which are higher than these of the pure metals.

(iii) They show conductivity like the of the pure metals.

(iv) They aquire chemical inertness.

Option (d) does not contain a pair of mixed oxides. Fe₂O₃ is not mixed oxide.

While, Mn₃O₄ (2MnO + MnO₂); Co₃O₄ (CoO + Co₂O₃); Fe₃O₄ (Fe₃O₃ + FeO) and Pb₃O₄ (2PbO + PbO₂), all are mixed salts.